Is ${183573}$ divisible by $9$ ?
Explanation: A number is divisible by $9$ if the sum of its digits is divisible by $9$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {183573}= &&{1}\cdot100000+ \\&&{8}\cdot10000+ \\&&{3}\cdot1000+ \\&&{5}\cdot100+ \\&&{7}\cdot10+ \\&&{3}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {183573}= &&{1}(99999+1)+ \\&&{8}(9999+1)+ \\&&{3}(999+1)+ \\&&{5}(99+1)+ \\&&{7}(9+1)+ \\&&{3} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {183573}= &&\gray{1\cdot99999}+ \\&&\gray{8\cdot9999}+ \\&&\gray{3\cdot999}+ \\&&\gray{5\cdot99}+ \\&&\gray{7\cdot9}+ \\&& {1}+{8}+{3}+{5}+{7}+{3} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $9$ , so the first five terms must all be multiples of $9$ That means that to figure out whether the original number is divisible by $9 $ , all we need to do is add up the digits and see if the sum is divisible by $9$ . In other words, ${183573}$ is divisible by $9$ if ${ 1}+{8}+{3}+{5}+{7}+{3}$ is divisible by $9$ Add the digits of ${183573}$ $ {1}+{8}+{3}+{5}+{7}+{3} = {27} $ If ${27}$ is divisible by $9$ , then ${183573}$ must also be divisible by $9$ ${27}$ is divisible by $9$, therefore ${183573}$ must also be divisible by $9$.